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I want to understand the proof given on the wikipedia page for the Wiener Algebra $A(\mathbb{T})$ (see here) that $A(\mathbb{T})$ is closed under multiplication. We have $$ A(\mathbb{T}) = \{f:\mathbb{T} \to \mathbb{R}: \|f\| < \infty \} $$ with norm $\| \cdot \|$ given by $$ \| f\| = \sum_{n\in \mathbb{Z}} |\hat f(n)|,$$ where $\hat f(n)$ is the $n$-th Fourier coefficient of $f$. The proof given on the wikipedia page is as follows: For $f,g\in A(\mathbb{T})$, we have \begin{align} f(t)g(t) & = \sum_{m\in\mathbb{Z}} \hat{f}(m)e^{imt}\,\cdot\,\sum_{n\in\mathbb{Z}} \hat{g}(n)e^{int} \\ & = \sum_{n,m\in\mathbb{Z}} \hat{f}(m)\hat{g}(n)e^{i(m+n)t} \\ & = \sum_{n\in\mathbb{Z}} \left\{ \sum_{m \in \mathbb{Z}} \hat{f}(n-m)\hat{g}(m) \right\}e^{int} . \end{align} Therefore, \begin{align} \|fg\| &= \sum_{n\in\mathbb{Z}} \left| \sum_{m \in \mathbb{Z}} \hat{f}(n-m)\hat{g}(m) \right| \\ &\leq \sum_{m} |\hat{f}(m)| \sum_n |\hat{g}(n)| \qquad\qquad (*)\\ &= \|f\| \, \|g\| \end{align}

The line that I don't understand is indicated by $(*)$. It is not clear to me why we can say that $$ \sum_{n\in\mathbb{Z}} \left| \sum_{m \in \mathbb{Z}} \hat{f}(n-m)\hat{g}(m) \right| \leq \sum_{m} |\hat{f}(m)| \sum_n |\hat{g}(n)|. $$ Why does this inequality hold? Any help is much appreciated.

Edit: Many thanks to @almosteverywhere for their answer. Using the discrete version of Young's convolution inequality, we have \begin{align} \sum_{n\in\mathbb{Z}} \left| \sum_{m \in \mathbb{Z}} \hat{f}(n-m)\hat{g}(m) \right| = \sum_{n\in\mathbb{Z}} \left| \widehat{f * g}(n) \right| = \| \widehat{f * g} \|_{\ell^1} \leq \| \hat{f} \|_{\ell^1}\| \hat{g} \|_{\ell^1} = \sum_{m\in \mathbb{Z}} |\hat{f}(m)| \sum_{n\in \mathbb{Z}} |\hat{g}(n)|. \end{align}

Martin Sleziak
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Matt E.
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2 Answers2

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For a simple proof: $$\sum_{n \in \mathbb Z} \Big| \sum_{m \in \mathbb Z} \hat f(n-m) \hat g(m) \Big | \leq \sum_{n \in \mathbb Z} \sum_{m \in \mathbb Z} | \hat f(n-m) | . | \hat g(m) | = \sum_{m \in \mathbb Z} | \hat g(m) | \sum_{n \in \mathbb Z} | \hat f(n-m) | $$ where you just need to switch the order of summation for the last part

roger
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They have used a discrete version of Young's convolution inequality: https://en.wikipedia.org/wiki/Young%27s_convolution_inequality#Generalizations.

More precisely, given an (locally compact) abelian group (i.e. $\mathbb{Z},+$) equipped with its bi-invariant Haar measure (i.e. the counting measure on $\mathbb{Z}$), we have $$ \lVert f* g \rVert_{L^1(\mathbb{Z})} \leq \lVert f \rVert_{L^1(\mathbb{Z})} \lVert g \rVert_{L^1(\mathbb{Z})} \, , $$ with the convolution defined the way you would expect it to be and integrable functions $f,g: \mathbb{Z}\to \mathbb{R}$.

almosteverywhere
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